Problem: Solve for $q$, $ -\dfrac{3q - 8}{q - 4} = -\dfrac{7}{3q - 12} - \dfrac{6}{4q - 16} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $q - 4$ $3q - 12$ and $4q - 16$ The common denominator is $12q - 48$ To get $12q - 48$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ -\dfrac{3q - 8}{q - 4} \times \dfrac{12}{12} = -\dfrac{36q - 96}{12q - 48} $ To get $12q - 48$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{7}{3q - 12} \times \dfrac{4}{4} = -\dfrac{28}{12q - 48} $ To get $12q - 48$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{6}{4q - 16} \times \dfrac{3}{3} = -\dfrac{18}{12q - 48} $ This give us: $ -\dfrac{36q - 96}{12q - 48} = -\dfrac{28}{12q - 48} - \dfrac{18}{12q - 48} $ If we multiply both sides of the equation by $12q - 48$ , we get: $ -36q + 96 = -28 - 18$ $ -36q + 96 = -46$ $ -36q = -142 $ $ q = \dfrac{71}{18}$